Solve (2-1)^2+2(z-1)(2z+3)+(2z+3)^2

$$( 2 - 1 ) ^ { 2 } + 2 ( z - 1 ) ( 2 z + 3 ) + ( 2 z + 3 ) ^ { 2 }$$

Evaluate

$$8z^{2}+14z+4$$

Solution Steps

Subtract 1 from 2 to get 1.

$$1^{2}+2\left(z-1\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$1+2\left(z-1\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Use the distributive property to multiply 2 by z-1.

$$1+\left(2z-2\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Use the distributive property to multiply 2z-2 by 2z+3 and combine like terms.

$$1+4z^{2}+2z-6+\left(2z+3\right)^{2}$$

Subtract 6 from 1 to get -5.

$$-5+4z^{2}+2z+\left(2z+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2z+3\right)^{2}.

$$-5+4z^{2}+2z+4z^{2}+12z+9$$

Combine 4z^{2} and 4z^{2} to get 8z^{2}.

$$-5+8z^{2}+2z+12z+9$$

Combine 2z and 12z to get 14z.

$$-5+8z^{2}+14z+9$$

Add -5 and 9 to get 4.

$$4+8z^{2}+14z$$

Expand

$$8z^{2}+14z+4$$

Solution Steps

Subtract 1 from 2 to get 1.

$$1^{2}+2\left(z-1\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$1+2\left(z-1\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Use the distributive property to multiply 2 by z-1.

$$1+\left(2z-2\right)\left(2z+3\right)+\left(2z+3\right)^{2}$$

Use the distributive property to multiply 2z-2 by 2z+3 and combine like terms.

$$1+4z^{2}+2z-6+\left(2z+3\right)^{2}$$

Subtract 6 from 1 to get -5.

$$-5+4z^{2}+2z+\left(2z+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2z+3\right)^{2}.

$$-5+4z^{2}+2z+4z^{2}+12z+9$$

Combine 4z^{2} and 4z^{2} to get 8z^{2}.

$$-5+8z^{2}+2z+12z+9$$

Combine 2z and 12z to get 14z.

$$-5+8z^{2}+14z+9$$

Add -5 and 9 to get 4.

$$4+8z^{2}+14z$$